16 $$ \begin{array}{l}\frac{3}{5} ext { of the pupils in Primary } 6 A ext { and } \frac{1}{2} ext { of the pupils in Primary } 6 B ext { are girls. } \ ext { Both classes have the same number of girls. Primary } 6 B ext { has } 6 ext { move boys } \ ext { than Primary } 6 A ext {. } \ ext { a) How many pupils are there in Primary } 6 B ext { ? } \ ext { b) } \frac{1}{3} ext { of the girls and } \frac{2}{5} ext { of the boys in the two classes wear spectacles. } \ ext { How many pupils in the two classes wear spectacles altogether? }\end{array} $$

Question

16 $$ \begin{array}{l}\frac{3}{5} 	ext { of the pupils in Primary } 6 A 	ext { and } \frac{1}{2} 	ext { of the pupils in Primary } 6 B 	ext { are girls. } \ 	ext { Both classes have the same number of girls. Primary } 6 B 	ext { has } 6 	ext { move boys } \ 	ext { than Primary } 6 A 	ext {. } \ 	ext { a) How many pupils are there in Primary } 6 B 	ext { ? } \ 	ext { b) } \frac{1}{3} 	ext { of the girls and } \frac{2}{5} 	ext { of the boys in the two classes wear spectacles. } \ 	ext { How many pupils in the two classes wear spectacles altogether? }\end{array} $$

UpStudy AI · Accepted Answer

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### **Given:**
1. **Primary 6 A:**
   - $$ \frac{3}{5} $$ of the pupils are girls.
2. **Primary 6 B:**
   - $$ \frac{1}{2} $$ of the pupils are girls.
3. Both classes have the **same number of girls**.
4. **Primary 6 B** has **6 more boys** than **Primary 6 A**.

### **Let’s Define Variables:**
- Let $$ G $$ be the number of girls in each class.
- Let $$ A $$ be the total number of pupils in Primary 6 A.
- Let $$ B $$ be the total number of pupils in Primary 6 B.

### **Using the Information:**
1. **For Primary 6 A:**
   $$
   \frac{3}{5}A = G \quad \Rightarrow \quad \text{Number of boys in 6 A} = A - G = A - \frac{3}{5}A = \frac{2}{5}A
   $$
   
2. **For Primary 6 B:**
   $$
   \frac{1}{2}B = G \quad \Rightarrow \quad \text{Number of boys in 6 B} = B - G = B - \frac{1}{2}B = \frac{1}{2}B
   $$
   
3. **Equal Number of Girls:**
   $$
   \frac{3}{5}A = \frac{1}{2}B \quad \Rightarrow \quad B = \frac{6}{5}A
   $$
   
4. **Difference in Number of Boys:**
   $$
   \frac{1}{2}B - \frac{2}{5}A = 6
   $$
   Substitute $$ B = \frac{6}{5}A $$:
   $$
   \frac{1}{2} \times \frac{6}{5}A - \frac{2}{5}A = 6 \quad \Rightarrow \quad \frac{3}{5}A - \frac{2}{5}A = 6 \quad \Rightarrow \quad \frac{1}{5}A = 6 \quad \Rightarrow \quad A = 30
   $$
   
5. **Total Pupils in Primary 6 B:**
   $$
   B = \frac{6}{5} \times 30 = 36
   $$

### **Answer to Part (a):**
**Primary 6 B** has **36 pupils**.

### **For Part (b):**
1. **Total Girls in Both Classes:**
   $$
   G_{\text{total}} = 18 + 18 = 36 \quad \left(\text{since } \frac{3}{5} \times 30 = 18 \text{ and } \frac{1}{2} \times 36 = 18\right)
   $$
   
2. **Total Boys in Both Classes:**
   $$
   B_{\text{total}} = 12 + 18 = 30 \quad \left(\text{since } \frac{2}{5} \times 30 = 12 \text{ and } \frac{1}{2} \times 36 = 18\right)
   $$
   
3. **Pupils Wearing Spectacles:**
   - $$ \frac{1}{3} $$ of the girls:
     $$
     \frac{1}{3} \times 36 = 12
     $$
   - $$ \frac{2}{5} $$ of the boys:
     $$
     \frac{2}{5} \times 30 = 12
     $$
   - **Total pupils wearing spectacles:**
     $$
     12 + 12 = 24
     $$

### **Answer to Part (b):**
A total of **24 pupils** in the two classes wear spectacles.

## **Final Answers:**
a) **36 pupils** in Primary 6 B.

b) **24 pupils** wear spectacles altogether.
**Primary 6 B has 36 pupils.** **A total of 24 pupils wear spectacles in both classes.**

16

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